A car travles 75 miles on 3 gallons of gasoline how much gasoline will it need to go 342 miles

Write the equation in standard form for the circle with center (5,0) passing through (-1, 9/2)

andreev551 [17]

Answer:

(x - 5)^2 + y^2 = 225/4,

or you could write it as (x - 5)^2 + y^2 = 56.25.

Step-by-step explanation:

The factor form is

(x - h)^2 + (y - k)^2 = r^2 where the center is (h, k) and r = the radius.

So we have:

(x - 5)^2 + (y - 0)^2 = r^2

As the point (-1, 9/2) is on the line:

(-1 - 5)^2 + (9/2)^2 = r^2

r^2 = 36 + 81/4

r^2 = 225/4.

So substituting for r^2:

(x - 5)^2 + (y - 0)^2 = 225/4

(x - 5)^2 + y^2 = 225/4 is the standard form.

3 0

3 years ago

Math is killing me right now. Help.

creativ13 [48]

Answer:

it 85 to the third power

Step-by-step explanation:

i gotchu gng

4 0

1 year ago

Read 2 more answers

if the volume of a rectangular prism is 560 cubic units, what is it's volume if one dimension is halved, a second dimension is r

Margaret [11]

70 cubic units.
___________

5 0

3 years ago

I don’t know what property to use

tiny-mole [99]

The AWSER is 12 I think

6 0

3 years ago

The overhead reach distances of adult females are normally distributed with a mean of 205 cm and a standard deviation of 7.8 cm.

devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z $\leq 1.72$ ). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15) from 1. Therefore, we have 1- P(Z $\leq -1.09$ ). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

4 0

3 years ago